**Problem of the Week**

Problem C and Solution

Around the Farm

**Problem**

Rahul has a farm he wishes to fence. The farm is the pentagon \(ABCDE\) shown below. He knows that \(ABCD\) is a 140 m by 150 m rectangle. He also knows that \(E\) is 50 m from the side \(AB\) and 30 m from the side \(BC\).

Determine the length of \(AE\), the length of \(DE\), and the perimeter of pentagon \(ABCDE\).

NOTE: The *Pythagorean Theorem* states, “In a right-angled triangle, the square of the length of hypotenuse (the side opposite the right angle) equals the sum of the squares of the lengths of the other two sides”.

In the following right triangle, \(p^2=r^2+q^2\).

**Solution**

Let \(F\) be the point on \(AB\) with \(EF=50\) m.

Let \(H\) be the point on \(BC\) with \(EH=30\) m.

Extend \(EF\) to \(G\) on \(CD\).

Since \(ABCD\) is a rectangle and \(FG\) is perpendicular to \(AB\), then \(FG\) is perpendicular to \(CD\) and \(FGCB\) is a rectangle. Therefore, \(FB= EH=GC=30\) m.

Also, \(DG = AF = AB - FB = 150 - 30 = 120\) m.

Since \(\triangle AFE\) and \(\triangle DGE\) are right-angled triangles, we can use the Pythagorean Theorem to determine the lengths of \(AE\) and \(DE\).

In \(\triangle AFE\), \[\begin{aligned} AE^2&=&AF^2+FE^2\\ &=&120^2+50^2\\ &=&14\,400+2500\\ &=&16\,900\\ AE&=&130\mbox{, since $AE > 0$}\\\end{aligned}\]

In \(\triangle DGE\), \[\begin{aligned} DE^2&=&DG^2+EG^2\\ &=&120^2+90^2\\ &=&14\,400+8100\\ &=&22\,500\\ DE&=&150\mbox{, since $DE > 0$}\end{aligned}\]

Therefore, \(AE=130\) m and \(DE=150\) m.

Also, the perimeter of pentagon \(ABCDE\) is equal to \(AB + BC + CD + DE + AE = 150 + 140 + 150 + 150 + 130= 720\) m.